KCET · Chemistry · Solid State
A metal crystallises in a body centred cubic lattice with the metallic radius \(\sqrt{3} \AA\). The volume of the unit cell in \(\mathrm{m}^3\) is
- A \(64 \times 10^{-29}\)
- B \(4 \times 10^{-29}\)
- C \(6.4 \times 10^{-29}\)
- D \(4 \times 10^{-10}\)
Answer & Solution
Correct Answer
(C) \(6.4 \times 10^{-29}\)
Step-by-step Solution
Detailed explanation
\(r=\sqrt{3} \AA=\sqrt{3} \times 10^{-10} \mathrm{~m}\)
For bcc,
\(\begin{aligned} & r=\frac{\sqrt{3} a}{4} \\ & a=\frac{4 r}{\sqrt{3}}=\frac{4 \times \sqrt{3} \times 10^{-10}}{\sqrt{3}}=4 \times 10^{-10} \\ & a^3=64 \times 10^{-30}=6.4 \times 10^{-29} \mathrm{~m}^3\end{aligned}\)
Thus, the volume of the unit cell is \(6.4 \times 10^{-29} \mathrm{~m}^3\).
For bcc,
\(\begin{aligned} & r=\frac{\sqrt{3} a}{4} \\ & a=\frac{4 r}{\sqrt{3}}=\frac{4 \times \sqrt{3} \times 10^{-10}}{\sqrt{3}}=4 \times 10^{-10} \\ & a^3=64 \times 10^{-30}=6.4 \times 10^{-29} \mathrm{~m}^3\end{aligned}\)
Thus, the volume of the unit cell is \(6.4 \times 10^{-29} \mathrm{~m}^3\).
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